2018 AIME II Problem 11

Below is the professionally curated solution for Problem 11 of the 2018 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AIME II solutions, or check the answer key.

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Concepts:permutationscomplementary countingrecursive counting

Difficulty rating: 3060

11.

Find the number of permutations of 1,2,3,4,5,61, 2, 3, 4, 5, 6 such that for each kk with 1k5,1 \le k \le 5, at least one of the first kk terms of the permutation is greater than k.k.

Solution:

The condition fails exactly when the first kk terms are a permutation of {1,,k}\{1, \ldots, k\} for some k5.k \le 5. For a permutation of 1,,n,1, \ldots, n, let kk be the smallest length for which the prefix is {1,,k}\{1, \ldots, k\} (the full length nn always works), and let cnc_n be the number of permutations whose smallest such kk is n.n. We want c6.c_6.

Every permutation of 1,,n1, \ldots, n decomposes uniquely as a minimal prefix of length kk (ckc_k choices) followed by any arrangement of the remaining nkn - k values, so k=1nck(nk)!=n!.\sum_{k=1}^{n} c_k \,(n-k)! = n!. Starting from c1=1,c_1 = 1, this gives c2=1,c_2 = 1, c3=3,c_3 = 3, c4=13,c_4 = 13, c5=71,c_5 = 71, and c6=720(1201+241+63+213+171)=720259=461.c_6 = 720 - (120 \cdot 1 + 24 \cdot 1 + 6 \cdot 3 + 2 \cdot 13 + 1 \cdot 71) = 720 - 259 = 461.

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